<p>A 45 g golf ball is pushed up a 30&#176; ramp. &#160;The ball starts from res

Question

<p>A 45 g golf ball is pushed up a 30&#176; ramp. &#160;The ball starts from rest and the length of the&#160;surface of the ramp is 5 m. &#160;The coefficient of kinetic friction between the surface on the ramp&#160;and the ball is 0.1. &#160;The force ceases when the ball leaves the top of the ramp and the total&#160;horizontal distance travelled by the ball after leaving the ramp is 20 m. &#160;What was the magnitude&#160;of the pushing force (in N)?</p>

Explanation / Answer

Let initial force = F

a = F/m - gcos(30) - gsin(30)

v^2 = 2a*5

Now,

time taken to fall = sqrt(2*2.5/9.81) = 0.714 sec.

so, v*0.714 = 20

v = 28.01

v^2 = 10a = 5.29

a = 0.529

F/m = 6.28

F = 0.283 N

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